WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects

Rectangular parallelopiped

In the length, breadth and height ofa rectangular parallelopiped be a unit, b unit and c unit respectively,then

Total surface area of the rectangular parallelopiped

= 2 x [Length x Breadth + Breadth x Height + Height x Length] sq-unit

Total surface area of the rectangular parallelepiped = 2 x (ab + be + ca) sq-unit.

Length of the diagonal of the rectangular

parallelopiped = \(\sqrt{(\text { Length })^2+(\text { Breadth })^2+(\text { Height })^2}\)unit

= \(\sqrt{a^2+b^2+c^2}\)unit

WBBSE Solutions for Class 10 Maths

Volume of the rectangular parallelopiped

= (Area of the base) x Height cubic-unit

= Length x Breadth x Height cubic unit

Volume of the rectangular parallelepiped = abc cubic-unit.

Cube:

If each side of a cube be a unit, then

The total surface area of the cube

= 6 x (side)² sq-unit

The total surface area of the cube= 6a² sq-unit

Length of the diagonal of the cube

= √3 x (Length of side) unit

Length of the diagonal of the cube = √3 a unit

Volume of the cube = (side)³ cubic-unit

Volume of the cube = a³ cubic-unit.

Right circular cylinder

If the radius of the base of a right circular cylinder be r unit and its height be h unit, then

Curved surface area of the cylinder

= (circumference of the base) x height sq-unit

= 2πr x h sq-unit

Curved surface area of the cylinder = 2πrh sq-unit

Total surface area of the cylinder

= (curved surface area) + (area of two end-circle) sq-unit

= (2πrh + 2πr²) sq-unit

Total surface area of the cylinder = 27π(h + r) sq-unit

Volume of the cylinder

= (Area of the circular base) x height cubic-unit

= πr² x h cubic-unit

Volume of the cylinder = πr²h cubic-unit.

Right circular cone:

If the radius of the base be r unit, height be h unit and slant height be l unit, then

Curved surface area of the cone

= \(\frac{1}{2}\) x (circumference of the base) x slant height sq-unit

= \(\frac{1}{2}\) x 2πr x l sq-unit

Curved surface area of the cone = πrl sq-unit.

Total surface area of the cone

= (curved surface area) + (area of the base) sq-unit

= (πrl +πr²) sq-unit

Total surface area of the cone = πr (l + r) sq-unit.

Volume of the cone

= \(\frac{1}{3}\) x (area of the base) x height cubic-unit

= \(\frac{1}{3}\) x (πr²) x h cubic-unit

Volume of the cone = \(\frac{1}{3}\) πr²h cubic-unit

Slant height of the cone

= \(\sqrt{(\text { radius of base })^2+(\text { height })^2}\) unit

Slant height of the cone = \(\sqrt{r^2+h^2}\) unit.

Sphere:

If the radius of the sphere be r unit, then

Total surface area of the sphere

= π x (diameter)² sq-unit

= π x (2r)² sq-unit

= π x 4r² sq-unit

Total surface area of the sphere = 4πr² sq-unit

Volume of the sphere

= \(\frac{4}{3}\) x π x (Radius)³ cubic-unit

Volume of the sphere = \(\frac{4}{3}\)πr³ cubic-unit.

Hemisphere

If the radius of a hemisphere, be r unit, then

Total surface area of a hollow hemisphere = 2πr² sq-unit

Curved surface area of a solid hemisphere = 2πr² sq-unit

Total surface area of a solid hemisphere = 3πr² sq-unit

Volume of a solid or hollow hemisphere = \(\frac{2}{3}\)πr³ cubic-unit

Mensuration Chapter 5 Problems Related To Different Solids And Objects Multiple Choice Questions

“WBBSE Class 10 problems on different solids solved examples”

Example 1. If a solid right circular cone of height r unit is made by melting a solid sphere of radius r unit, then the radius of the base of the cone will be

1. 2r unit
2. 3r unit
3. r unit
4. 4r unit

Solution:

Given:

If a solid right circular cone of height r unit is made by melting a solid sphere of radius r unit

Radius of the sphere = r unit

∴ Volume of the sphere = \(\frac{4}{3}\) πr³ cubic-unit.

Also, the height of the cone = r unit

Let the radius of the base of the cone be r₁ unit

∴ Let the radius of the base = r₁ unit

∴ the volume of the cone = \(=\frac{1}{3} \pi r_1^2\) x r cubic-unit

As per question, \(\frac{1}{3} \pi r_1^2 \times r=\frac{4}{3} \pi r^3\)

⇒ \(r_1^2=4 r^2\)

⇒ \(r_1^2=(2 r)^2\)

⇒ r₁ = 2r

∴ radius of the cone = 2r unit

Hence 1. 2r unit is correct

Example 2. The radius of a right circular cylinder is r un it and height is 2r units. The diameter of the largest sphere that can be put into the cylinder will be

1. r unit
2. 2r unit
3. \(\frac{r}{2}\)
4. 4r unit

Solution:

Given:

The radius of a right circular cylinder is r un it and height is 2r units.

The radius of the cylinder = r unit and height = 2r unit.

∴ the diameter of the sphere = 2r unit.

Hence 2. 2r unit is correct.

Example 3. The volume of the largest solid cone that can be cut off from a solid hemisphere will be

1. 4πr³ cubic-unit
2. 3πr³ cubic-unit
3. \(\frac{\pi r^3}{4}\) cubic-unit
4. \(\frac{\pi r^3}{3}\) cubic-unit

Solution:

Given:

The volume of the largest solid cone that can be cut off from a solid hemisphere

The radius of the hemisphere = r unit

∴ the radius of the cone = r unit

and height of the cone = r unit

∴ volume of the cone = \(\frac{1}{3}\)πr² x r cubic-unit

= \(\frac{\pi r^3}{3}\) cubic unit

Hence 4. \(\frac{\pi r^3}{3}\) cubic-unit is correct.

Example 4. The radius of the largest sphere that can be cut off from a solid cube of side x unit each will be

1. x unit
2. 2x unit
3. \(\frac{x}{2}\) unit
4. 4x unit

Solution:

Given:

The radius of the largest sphere that can be cut off from a solid cube of side x unit

The length of each side of the cube = x unit

∴ the diameter of the largest sphere = x unit

∴ the radius of the largest sphere = \(\frac{x}{2}\) unit

Hence 3. \(\frac{x}{2}\) unit is correct.

“Mensuration exercises for different solids Class 10”

Example 5. The height of the water level when the water of a right circular cone-shaped bottle of radius r unit and of height h unit is poured into a right circular cylinder-shaped pot of radius mr unit will be

1. \(\frac{h}{2} m^2\) unit
2. \(\frac{2 h}{m}\) unit
3. \(\frac{h}{3 m^2}\) unit
4. \(\frac{m}{2 h}\) unit

Solution:

Given:

The height of the water level when the water of a right circular cone-shaped bottle of radius r unit and of height h unit is poured into a right circular cylinder-shaped pot

The radius of the conical bottle = r unit and height = h unit

∴ the volume of the bottle = \(\frac{1}{3}\) πr²h cubic-unit.

Again, the radius of the cylindrical pot = mr unit,

Let the water-level will rise x unit.

∴ the volume of the water poured into the pot = π x (mr)² x x cubic-unit

∴ π(mr)²x = \(\frac{1}{3}\)h

⇒ m²r² x = \(\frac{r^2 h}{3}\)

⇒ x = \(\frac{h}{3 m^2}\)

the water-level of the pot will rise unit.

Hence 3. \(\frac{h}{3 m^2}\) unit is correct.

Mensuration Chapter 5 Problems Related To Different Solids And Objects True Or False

Example 1. If two solid hemispheres of same type of which the radii of bases are r unit be joined along their bases, then the total surface area of the joint solid object is 6πr² square-unit.

Solution: False

Since the radii of the hemispheres is r unit and the two hemispheres are joint along their bases, so we shall get a full sphere, the total surface area which will be 4πr²sq-unit

Example 2. The radius of base, height and slant height of a solid right circular cone are r unit, h unit and l unit respectively. The base of the cone is joint along the base of solid right circular cylinder. If the radii of the bases and heights of the cylinder and of the cone be equal, then the total surface area of the joint solid object will be (πrl + 2πrh + 2πr²) sq-unit.

Solution: False.

Since the radius of the base of the cone is r unit and its slant height is l unit.

∴ curved surface area = πrl sq-unit.

Again, the radius of the cylinder is r unit and height is h unit.

∴ the surface area of the solid object = (πrl + 2πrh + πr²) sq-unit.

Mensuration Chapter 5 Problems Related To Different Solids And Objects Fill In The Blanks

Example 1. The radii of bases of a solid right circular cone and of two hemispheres are equal. If the two hemispheres be joined with two end-planes of the cone, then the total surface area of the new solid object = (the curved surface area of one of the hemispheres) + (curved surface area of the ______ ) + (the curved surface area of the other hemisphere).

Solution: cylinder

Example 2. The shape of a pencil with its one end cut is the coordination of a right circular cone and a right circular ______

Solution: cylinder

Example 3. If a solid sphere is melted to make a solid right circular cylinder, then the volumes of the sphere and the cylinder are ______

Solution: equal

Mensuration Chapter 5 Problems Related To Different Solids And Objects Short Answer Type Questions

“Chapter 5 different solids problems WBBSE solutions”

Example 1. If a solid sphere of radius r unit be melted to make a solid right circular cone of height r unit, then find the radius of the base of the cone.

Solution:

Given:

A solid sphere of radius r unit be melted to make a solid right circular cone of height r unit

The radius of the sphere = r unit.

∴ the volume of the sphere = \(\frac{4}{3}\)πr³ cubic-unit.

Again, the height of the cone = r unit

Let the radius of the cone = x unit.

∴ the volume of the cone = \(\frac{1}{3}\)πx²r cubic-unit.

As per question, \(\frac{1}{3}\)πx²r = \(\frac{4}{3}\)πr³

⇒ x² = 4r²

⇒ x² = (2r)²  ⇒ x = 2r

Hence the radius of base of the cone is 2r unit.

Example 2. If the radii and volumes of a solid right circular cone and a solid sphere be equal, then what will be the ratio of the length of diameter of the sphere and the height of the cone?

Solution:

Given:

The radii and volumes of a solid right circular cone and a solid sphere be equal

Let the radii of the cone and the sphere be r unit and the height of the cone be x-unit.

∴ the volume of the cone = \(\frac{1}{3}\)πr²x cubic-uinit

and the volume of the sphere = \(\frac{4}{3}\)πr³ cubic-unit.

As per question,\(\frac{4}{3}\)πr³ = \(\frac{1}{3}\)πr²x

⇒ 4r = x

⇒ 2r = \(\frac{x}{2}\) = \(\frac{1}{2}\)

⇒ 2r: x = 1 : 2

Hence. the required ratio is 1:2.

Example 3. The shape of the lower part of a solid object is like a hemisphere and the upper part like a right circular cone. If the surface area of the two parts be equal, then find the ratio of the length of radius and height of the cone.

Solution:

Given:

The shape of the lower part of a solid object is like a hemisphere and the upper part like a right circular cone. If the surface area of the two parts be equal

Let the radius of the hemisphere = r unit

∴ the radius of the cone = r unit

Let the height of the cone = h unit.

As per the question,

27πr² = πrl, where l = slant height of the cone.

⇒ 2r = l ⇒ 2r = \(\sqrt{h^2+r^2}\)

⇒ 4r² = h² + r²  ⇒ 3r² = h²

⇒ √3r = h

⇒ \(\frac{r}{h}=\frac{1}{\sqrt{3}}\)

⇒ r: h = 1: √3

Hence the required ratio = 1 : √3.

“Class 10 Maths mensuration problems on various objects”

Example 4. The length ofradius of base of a solid right circular cone is equal to the length of radius of a solid sphere. If the volume of the sphere is 2 times the volume of the cone, then find the ratio of the height and radius of base of the cone.

Solution:

Given:

The length ofradius of base of a solid right circular cone is equal to the length of radius of a solid sphere. If the volume of the sphere is 2 times the volume of the cone

Let the radius of the base of the cone = r unit

∴ the radius of the sphere = r unit

Let the height of the cone = h unit

∴ the volume of the cone = \(\frac{1}{3}\)πr²h cubic-uinit

and the volume of the sphere = \(\frac{4}{3}\)πr³ cubic-unit.

As per question, 2 x \(\frac{1}{3}\)πr²h = \(\frac{4}{3}\)πr³

⇒ \(\frac{h}{r}=\frac{4}{2} \Rightarrow \frac{h}{r}=\frac{2}{1}\)

⇒h: r = 2: 1

Hence the required ratio is 2: 1.

Example 5. If a right circular cone of height 15 cm and of diameter 30 cm is made from a wooden sphere of radius 15 cm, then what percent of wood will be lost?

Solution:

Given:

A right circular cone of height 15 cm and of diameter 30 cm is made from a wooden sphere of radius 15 cm

The radius of the sphere = 15 cm

∴ the volume of the sphere = \(\frac{4}{3}\)π x (15)³ cc.

Now, the radius of the cone = \(\frac{30}{2}\) cm= 15 cm [∵ diameter = 30 cm]

and the height of the cone =15

∴ the volume of the cone = \(\frac{1}{3}\)π x (15)² x 15 cc

= \(\frac{1}{3}\)π x (15)³ cc

∴ the lost wood = \(\left\{\frac{4}{3} \pi \cdot \times(15)^3-\frac{1}{3} \pi \times(15)^3\right\}\) cc

= \(\left(\frac{4}{3}-\frac{1}{3}\right) \pi \times(15)^3 \mathrm{cc}=\pi \times(15)^3 \mathrm{cc}\)

∴ the percent of lost wood = \(\frac{\pi \times(15)^3}{\frac{4}{3} \pi \times(15)^3}\) x 100% = 75%

Hence the required percentage = 75%.

Example 6. How many part of a rectangular parallelopiped type hole of length, breadth and depth 48 cm, 16.5 cm and 4 cm respectively should be stuffed with the soil obtained by digging a right circular conical tunnel of diameter 4 metres and of length 56 metres?

Solution: The volume of the soil obtained = π x \(\left(\frac{4}{2}\right)^2\) x 56 cubic-metre.

Again, the volume of the hole = 48 x 16.5 x 4 cubic-metre.

Let x part of the hole should be stuffed.

∴ x x 48 x 16.5 x 4 = π x 2² x 56

or, \(x=\frac{\frac{22}{7} \times 4 \times 56}{48 \times 16.5 \times 4}\)

or, x= \(\frac{22 \times 4 \times 8 \times 10}{48 \times 165 \times 4}\)

or, x = \(\frac{2}{9}\)

Hence \(\frac{2}{9}\) part of the hole should be stuffed.

“Understanding different solids in Class 10 Maths”

Example 7. What will be the ratio of the diameter of the sphere and the height of a cylinder when the volumes of the sphere and the right circular cylinder are equal, given that the radii of the sphere and the cone are equal?

Solution: Let the radii of the sphere and the cylinder be r unit and the height of the cylinder be h unit.

As per question, πr²h = \(\frac{4}{3}\)πr³

⇒ 3h = 4r ⇒ 2.2r = 3h

⇒ \(\frac{2 r}{h}=\frac{3}{2}\)

⇒ 2r: h =3: 2

Hence the,ratio of the diameter of the sphere and the height of the cylinder is 3: 2.

Example 8. The ratio of the radii of a right circular cylinder and a right circular cone is 3: 4 and the ratio of their heights is 2 : 3..Then find the ratio of the volumes of them.

Solution:

Given:

The ratio of the radii of a right circular cylinder and a right circular cone is 3: 4 and the ratio of their heights is 2 : 3..

Let the radii Of the cylinder and the cone be 3r unit and 4r unit respectively and their heights be 2h unit and 3h unit respectively.

∴ Volume of the cylinder = π x (3r)² x 2h cubic-unit = 18πr²h cubic-unit

and volume of the cone = \(\frac{1}{3}\)π x (4r)² x 3h cubic-unit = 16πr²h cubic-unit

Hence the ratio of the volumes of cylinder and the cone is 18πr²h: 16πr²h

= \(\frac{18 \pi r^2 h}{16 \pi r^2 h}=\frac{9}{8}\) = 9: 8

Hence the required ratio = 9:8.

Example 9. How many spherical bullets of diameter 4 cm should be got from a cube of edges 44 cm?

Solution: The volume of the cube = (44)³ cc

The diameter of each spherical bullet = 4 cm

∴ the radius of each spherical bullet = \(\frac{4}{2}\)cm = 2 cm

the volume of each spherical bullet = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 2³ cc

Let the number, of bullets to be got = x.

So, x = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 2³ = (44)³

⇒ x = \(\frac{44 \times 44 \times 44 \times 3 \times 7}{4 \times 22 \times 2 \times 2 \times 2}\) = 2541

Hence the number of required bullets = 2541.

“Step-by-step solutions for solids and objects Class 10”

Example 10. If the curved surface areas of a sphere and a right circular cylinder both of same radii be equal, then what will be the ratio of their volumes?

Solution:

Given:

The curved surface areas of a sphere and a right circular cylinder both of same radii be equal

Let the radii of both the sphere and the cylinder be r unit and also let the height of the cylinder be h unit.

∴ the curved surface area of the sphere = 4πr² sq-unit

and the curved surface area of the cylinder = 2πrh sq-unit.

As per the question, 4πr² = 2πrh

or, \(\frac{\pi r^2}{\pi r h}=\frac{2}{4} \quad \text { or, } \quad \frac{r}{h}=\frac{1}{2}\)

or, h = 2r

∴ the ratio of the volumes of the sphere and the cylinder

= \(\frac{4}{3} \pi r^3: \pi r^2 h\)

= \(\frac{\frac{4}{3} \pi r^3}{\pi r^2 h}=\frac{4 \pi r^3}{3 \pi r^2 \times 2 r} \cdot\) [∵ because h=2 r]

= \(\frac{4 \pi r^3}{6 \pi r^3}=\frac{4}{6}=\frac{2}{3}\) =2: 3

Hence the required ratio of the volumes of sphere and the cylinder = 2:3.

Mensuration Chapter 5 Problems Related To Different Solids And Objects Long Answer Type Questions

“Examples of problems related to solids for Class 10 Maths”

Example 1. There is a solid iron-pillar in front of Anil’s house, the lower part of which is of type of a right circular cylinder and the upper part is of type of a right circular cone. The radii of their diameter of bases is 20 cm. If the height of the cylindrical part be 2.8 metres and that of the conical part be 42 cm and if the weight of1 cc iron be 7.5 gm, then what will be the weight of the pillar?

Solution:

Given:

There is a solid iron-pillar in front of Anil’s house, the lower part of which is of type of a right circular cylinder and the upper part is of type of a right circular cone. The radii of their diameter of bases is 20 cm. If the height of the cylindrical part be 2.8 metres and that of the conical part be 42 cm and if the weight of1 cc iron be 7.5 gm

The diameter of the bases = 20 cm

∴ The radii of the bases = \(\frac{20}{2}\) cm = 10 cm

The height of the cylindrical part = 2.8 metres = 280 cm.

∴ the volume of this part= \(\frac{22}{7}\) x (10)² x 280 cc

= \(\frac{22}{7}\) x 100 x 280 cc

= 88000 cc

Again, the height of the conical part = 42 cm

∴ the volume of the conical part = \(\frac{1}{3}\) x \(\frac{22}{7}\) x (10)² x 42 cc

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 100 x 42 cc

= 4400 cc

So, the total volume of the pillar = (88000 + 4400) cc = 92400 cc

∴ the weight of the total pillar = (7.5 x 92400) gm = 693000 gm = 693 kg

Hence the required weight = 693 kg.

Example 2. There is some water in a right circular cylindrical pot of diameter 24 cm. How much the height of water-level will be increased if 60 solid conical iron piece of diameter 6 cm and of height 4 cm each are completely immersed into that water?

Solution:

Given:

There is some water in a right circular cylindrical pot of diameter 24 cm.

The diameter of the conical iron-piece = 6 cm

∴ the radius of the conical iron piece = \(\frac{6}{2}\) cm = 3 cm

Also, the heights of the iron-pieces = 4 cm

So, the volume of each iron-piece = \(\frac{1}{2}\) x π x (3)² x 4 cc = 12π cc

Again, the diameter of the conical pot = 24 cm

the radius of the conical pot = \(\frac{24}{2}\) cm = 12 cm

Let water-level will rise h cm

∴ the volume of the raised water= π x (12)² x h cc = 144 πh cc

As per the question, 144 πh = 60 x 12π

⇒ h = \(\frac{60 \times 12 \pi}{144 \pi}\)

⇒ h = 5

Hence the water-level of the pot will rise 5 cm.

Example 3. The ratio of the curved surface areas of a solid right circular cone and of a right circular cylinder both of same radius of base and of same height is 5: 8. Then find the ratio of their radius of base and height.

Solution:

Given:

The ratio of the curved surface areas of a solid right circular cone and of a right circular cylinder both of same radius of base and of same height is 5: 8.

Let radius of base = r unit

and height of them = h unit

Also, let the slant height of the cone = l unit.

∴ curved surface area of the cone = πrl sq-unit

= \(\pi r \sqrt{r^2+h^2}\) sq-unit

The curved surface area of the cylinder = 2πrh sq-unit.

As per question, \(\frac{\pi r \sqrt{r^2+h^2}}{2 \pi r h}=\frac{5}{8}\)

or, \(\frac{\sqrt{r^2+h^2}}{h}=\frac{5}{4}\)

or, \(\frac{r^2+h^2}{h^2}=\frac{25}{16}\)

or, \(\frac{r^2}{h^2}+1=\frac{25}{16}\)

or, \(\frac{r^2}{h^2}=\frac{25}{16}-1\)

or, \(\frac{r^2}{h^2}=\frac{9}{16}\)

or, \(\frac{r}{h}=\sqrt{\frac{9}{16}}=\frac{3}{4}\)

⇒ r: h = 3: 4

Hence the required ratio of the radius of base and the height = 3: 4

Example 4. The radius of cross-section of a right circular rod is 3.2 dcm. The rod is nielted to make 21 solid spheres. If the radius of each sphere be 8 cm, then what will be the length of the rod?

Solution:

Given:

The radius of cross-section of a right circular rod is 3.2 dcm. The rod is nielted to make 21 solid spheres. If the radius of each sphere be 8 cm

The radius of the right circular rod = 3.2 dcm = 32 cm.

Let the length of the rod =x cm

∴ the volume of the rod = π x (32)² x x cc

The radius of each spheres = 8 cm

∴ the volume of each sphere = \(\frac{4}{3}\) x π x (8)³ cc

So, the volume of 21 spheres = 21 x \(\frac{4}{3}\) x π x (8)³ cc

= 28 x 512 π cc

As per question, π x 32 x 32 x = 28 x 512 π

⇒ x = \(\frac{28 \times 512}{32 \times 32}\) = 14

Hence the required length of the rod = 14 cm.

Example 5. How many solid spheres of diameter 2.1 dcm each can be made by melting a solid rectangular parallelopiped copper piece of length 6.6 dcm, of breadth 4.2 dcm and of thickness 1.4 dcm? How many cubic-dcm metal will possess in each- sphere?

Solution: The length of copper piece = 6.6 dcm = 66 cm

The breadth of copper piece = 4.2 dcm = 42 cm

Height of copper piece= 1.4 dcm = 14 cm

So, the volume of coper piece = (66 x 42 x 14) cc

The diameter of each sphere = 2.1 dcm = 21 cm

∴ the radius of each sphere = \(\frac{21}{2}\) cm

∴ the volume of each sphere = \(\frac{4}{3}\) x \(\frac{22}{7}\) x \(\left(\frac{21}{2}\right)^3\) cc

= \(\frac{4}{3} \times \frac{22}{7} \times \frac{21 \times 21 \times 21}{2 \times 2 \times 2}\)

= 11 x 21 x 21 cc

Let the number of spheres be x.

∴ x x 11 x 21 x 21 = 66 x 42 x 14

⇒ x = \(\frac{66 \times 42 \times 14}{11 \times 21 \times 21}\)

⇒ x = 8

Also, the volume of each sphere = 11 x 21 x 21cc

= \(\frac{11 \times 21 \times 21}{1000}\) cubic-dcm = 4.851. cubic dcm.

Hence the required number of sphere is 8 and the volume of each sphere is 4.851 cubic-dcm.

Example 6. The radius of base of a solid right circular iron-rod is 32 cm and the length of the rod is 35 cm. How many solid right circular cone of radius of base 8 cm and of height 28 Cm can be made by melting this rod?

Solution:

Given:

The radius of base of a solid right circular iron-rod is 32 cm and the length of the rod is 35 cm.

The radius of the base of the rod = 32 cm

and its length = 35 cm.

∴ volume of the rod = π x (32)² x 35 Cc

Again, the radius of base of each cone = 8 cm

and height of each cone = 28 cm.

∴ volume = \(\frac{1}{3}\) x π x (8)² x 28 cc

Let the number of cone be x.

∴ \(\frac{1}{3}\)π x (8)² x 28 x x = π x (32)² x 35

⇒ x =\(x=\frac{32 \times 32 \times 35 \times 3}{8 \times 8 \times 28} .\)

⇒ x = 60.

Hence the required number of solid cone is 60.

“WBBSE Mensuration Chapter 5 practice questions on solids”

Example 7. The external radius of a hollow sphere of thickness 1 cm made of lead-plate is 6 cm. If a solid right circular rod of radius 2 cm is made by melting this hollow sphere, then what will be the length of the rod?

Solution:

Given:

The external radius of a hollow sphere of thickness 1 cm made of lead-plate is 6 cm. If a solid right circular rod of radius 2 cm is made by melting this hollow sphere

The external radius of the hollow sphere = 6 cm

The thickness of the sphere = 1 cm

∴ the inner radius of the hollow sphere = (6 – 1) cm = 5 cm

So, the volume of the materials of the sphere = \(\frac{4}{3}\)π(6³ – 5³) cc

= \(\frac{4}{3}\)π(216 -125) cc

= \(\frac{4}{3}\)π x 91 cc

The radius of the right circular rod = 2 cm.

Let the length of the rod = x cm

∴ the volume of the rod = π x (2)² x x cc = 4πx cc

As per question, 4πx= \(\frac{4}{3}\) x π x 91

⇒ x =\(\frac{4 \times 91}{3 \times 4}\)

⇒ x = \(\frac{91}{3}=30 \frac{1}{3}\)

Hence the length of the solid right circular rod =30\(\frac{1}{3}\) cm

Example 8. What will be the ratio of the volumes of a solid cone, a solid hemisphere and a solid cylinder when their radii of bases and heights are equal?

Solution:

Let radius of the cone, hemisphere and cylinder be r unit and their equal heights be h unit.

Since the heights of the three solid objects are equal,

∴ height of the cone = height of the cylinder = height of the hemisphere = radius of the hemisphere = r unit.

∴ volume of the cone: volume of the hemisphere: volume of the cylinder

= \(\frac{1}{3} \pi r^2 h: \frac{2}{3} \pi r^3: \pi r^2 h\)

= \(\frac{1}{3} \pi r^2 \cdot r: \frac{2}{3} \pi r^3 \cdot: \pi r^2 \cdot r\)

= \(\frac{1}{3} \pi r^3: \frac{2}{3} \pi r^3: \pi r^3\)

= \(\frac{1}{3}: \frac{2}{3}: 1\)

= 1:2 : 3.

Hence the required ratio is 1 : 2 : 3.

Example 9. The upper part of a solid right circular cylindrical pillar is a hemisphere. What will be the volume of the pillar if the radius of base of it is 2 metres and its total length is 10 metres?

Solution:

Given:

The upper part of a solid right circular cylindrical pillar is a hemisphere.

Since the radius of base is 2 metres, so the radius of the hemisphere of the upper, part of the pillar is 2 metres.

∴ the height of the cylindrical part = (10 – 2) metres = 8 metres.

Now, the volume of the hemisphere =\(\frac{1}{2}\) :\(\frac{4}{3}\) x π x 2³ cubic-metre

= \(\frac{16}{3}\)π cubic-metre

and the volume of the cylindrical part = π x (2)² x 8 cubic-metre = 32π cubic-metre.

∴ the total volume = \(\left(\frac{16}{3} \pi+32 \pi\right)\)cubic-metre

= \(\pi\left(\frac{16}{3}+32\right)\)cubic-metre

= \(\pi \times \frac{112}{3}\)cubic-metre

= \(\frac{22}{7} \times \frac{112}{3}\)cubic-metre

= \(\frac{352}{3}\)cubic-metre

=117\(\frac{1}{3}\)cubic-metres

Hence the total volume of the pillar 117\(\frac{1}{3}\) cubic-metres.

Example 10. The lower part of a tent is of shape of a right circular cylinder, the height of which is 3.5 metres. The upper part of the tent is of shape of a right circular cone. If the total height of the tent be 9.5 metre and its diameter of base be 5 metre, then how much quantity of tarpaulin will be required to make 14 such tents?

Solution:

Given:

The lower part of a tent is of shape of a right circular cylinder, the height of which is 3.5 metres. The upper part of the tent is of shape of a right circular cone. If the total height of the tent be 9.5 metre and its diameter of base be 5 metre

The diameter of the base of the tent = 5 cm

∴ the radius of the base of the tent = \(\frac{5}{2}\) = 2.5 m

height of the cylindrical part = 3.5 m

∴ the curved surface area of this part = 2πrh = 2 x\(\frac{22}{7}\) x 2.5 x 3.5 sq-metre = 55 sq-metre

The radius of the conical upper part of the tent = 2.5 m and height = (9.5 – 3.5) m = 6 m.

The slant height of the conical part = \(\sqrt{r^2+h^2}\)

or, \(l=\sqrt{(2.5)^2+(6)^2} \mathrm{~m}\)

or, \(l=\sqrt{42.25} \mathrm{~m}\) = 65 m

∴ the curved surface area of this conical part = πrl

= \(\frac{22}{7}\) x 25 x 6.5 sq-m

∴ the tarpaulin required to make one tent = \(\left(55+\frac{22}{7} \times 2.5 \times 6.5\right)\)sq-metre

∴ the tarpaulin required to make 14 tent = 14 \(\left(55+\frac{22}{7} \times 2.5 \times 6.5\right)\) x 2.5 x 65 = 1485 sq-metre.

Hence the required quantity of tarpaulin = 1485 sq-metre.

Example 11. A conical flask of height 24 cm is full of water. If this water is poured into another cylindrical flask of radius half of the conical flask, then how much water-level of the cylindrical flask will be raised?

Solution:

Given:

A conical flask of height 24 cm is full of water. If this water is poured into another cylindrical flask of radius half of the conical flask

Let the radius of the base of the conical flask = r cm

∴ the radius of base of the cylindrical flask = \(\frac{r}{2}\) cm

∴ the volume of the conical flask = \(\frac{1}{3}\)πr² cc

= \(\frac{1}{3}\)πr² x 24 cc = 8πr² cc

Let water-level into the cylindrical flask will rise h cm.

As per condition, π x \(\left(\frac{r}{2}\right)^2\) x h = 8πr²

\(\frac{1}{4}\)πr²h = 8πr² ⇒ h = 32.

Hence the water-level into the cylindrical flask will rise 32 cm.

Example 12. One part of an Iron-pillar is cylimlrical and the rest part is conical. The radii of the cylindrical and conical part arc 8 cm and their heights are 240 cm and 36 cm respectively. If the weight of I ec iron he 7.8 gm, then what will be the total weight of the pillar?

Solution:

Given:

One part of an Iron-pillar is cylimlrical and the rest part is conical. The radii of the cylindrical and conical part arc 8 cm and their heights are 240 cm and 36 cm respectively. If the weight of I ec iron he 7.8 gm

The radius of the cylindrical part 8 cm and its height = 240 cm.

∴ the volume of the cylindrical part = \(\frac{22}{7}\) x 8² x 240 cubic cm

= \(\frac{22}{7}\)x 64 x 240 cc

Again, the radius of the conical part = 8 cm and its height = 36 cm.

∴ the volume of the conical part \(\frac{1}{3}\) x \(\frac{22}{7}\) x 8² x 36 cc

= \(\frac{22}{7}\) x 64 x 12 cc

So, the total volume of the pillar = \(\left(\frac{22}{7} \times 64 \times 240+\frac{22}{7} \times 64 \times 12\right)\) cc

= \(\frac{22}{7}\) x 64 x (240 + 12) cc

= \(\frac{22}{7}\) x 64 x 252 cc

Now, the weight of 1 cc pillar is 7.8 gm.

∴ the weight of \(\frac{22}{7}\) x 64 x 252 cc is 7.8 x \(\frac{22}{7}\) x 64 x 252 gm

= 395366.4 gm = 395.3664 kg

Hence the weight of the total pillar = 395.37 kg (approx.)

Example 13. Water flows at a speed of 10 m per minute through a right circular pipe of diameter 5 mm. In how much time a tank of conical shape having diameter 40 cm and depth 24 cm will be fulfilled completely by this pipe?

Solution:

Given:

Water flows at a speed of 10 m per minute through a right circular pipe of diameter 5 mm.

The diameter of the conical tank = 40 cm

∴ its radius = \(\frac{40}{2}\) cm = 20 cm

The depth of the tank = 24 cm

∴ the volume of the tank = \(\frac{1}{3}\) x π x (20)² x 24 cc

= \(\frac{1}{3}\)π x x 20 x 20 x 24 cc = 3200π cc

Let the tank will be fulfilled completely in x minute.

∴ the length of the water-column = 10x metres = 1000x cm

Clearly, the length of the cylindrical water-column is 1000x cm

and its radius = \(\frac{5}{2}\) mm = \(=\frac{5}{2 \times 10}\) cm = \(\frac{1}{4}\) cm

∴ the volume of cylindrical water-column = \(\pi \times\left(\frac{1}{4}\right)^2 \times 1000 x \mathrm{cc}\)

= \(\pi \times \frac{1}{16} \times 1000 x \mathrm{cc}\)

As per question, \(\frac{\pi \times 1000 x}{16}\) = 3200π

⇒ x = \(\frac{3200 \times 16}{1000}=\frac{256}{5}=51 \frac{1}{5}\)

Hence the tank will be completely fulfilled in 51\(\frac{1}{5}\) minutes.

“WBBSE Class 10 Maths solved problems on mensuration of solids”

Example 14. A pot of conical shaped, the radius and height of which is 6 cm and 8 cm respectively, is fulfilled with water. Now, a sphere is completely immersed into the water of this pot in such a wayn that it just touches the two sides of the pot. So, how many part of the water of the pot will over flow?

Solution:

Given:

A pot of conical shaped, the radius and height of which is 6 cm and 8 cm respectively, is fulfilled with water. Now, a sphere is completely immersed into the water of this pot in such a wayn that it just touches the two sides of the pot.

AD = 6 cm and DC = 8 cm

∴ AC = \(\sqrt{C D^2+A D^2}\)

= \(\sqrt{8^2+6^2} \mathrm{~cm}\)

= \(\sqrt{64+36} \mathrm{~cm}\)

= 10 cm

Now, in the right-angled triangles ΔACD and ΔEOC, we get, ∠ACD = ∠ECO

Let the radius of the sphere be x cm.

∴ AD = AE = 6 cm

∴ EC = (AC – AE) = (10 – 6) cm = 4 cm

Again, ΔADC ~ ΔEOC,

∴ \(\frac{\mathrm{DC}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{EO}}\)

⇒ \(\frac{8}{6}=\frac{4}{x}\)

⇒ \(x=\frac{4 \times 6}{8}=3\)

∴ the radius of the sphere is 3 cm

So, the volume of the sphere = \(\frac{4}{3}\)π x 3³ cc = 3671 cc

Again, the volume of the conical pot = \(\frac{1}{3}\)π x 6² x 8 cc= 96π cc

So, the required part = \(\frac{36 \pi}{96 \pi}=\frac{3}{8}\)

Hence \(\frac{3}{8}\) part of the water of the pot will overflow.

Example 15. The lower part of a tent is a right circular cylinder of radius 14 metres and its upper part is a right circular cone. If the height of the cylindrical part is 3 metre and the total height of the tent is 13.5 metre, then find the expenditure of colouring the inner part of the tent at a rate of 2 per square-metre.

Solution:

Given:

The lower part of a tent is a right circular cylinder of radius 14 metres and its upper part is a right circular cone. If the height of the cylindrical part is 3 metre and the total height of the tent is 13.5 metre

The radius of the cylindrical part = 1 4 m and the height of this part = 3 m.

∴ the curved surface area of the cylindrical part = 2 x \(\frac{22}{7}\)x 14 x 3 square-metre = 264 square-metre

Again, the height of the conical part of the tent = (13.5 – 3) m = 10.5 m

and its radius of base = 14 metre

∴ the slant height of the conical part

= \(\sqrt{(14)^2+(10.5)^2}\) metre

= \(\sqrt{196+110.25}\) metre

= \(\sqrt{306.25}\) metre

= 17.5 metre

∴ the curved surface area of the conical part of the tent = \(\frac{22}{7}\) x 14 x 17.5 sq-metre = 770 square-metre

∴ the area of the part of the tent which is to be coloured = (770 + 264) square-metre = 1034 square-metre

So the expenditure of colouring = ₹2 x 1034 = ₹2068

Hence the required expenditure = ₹2068.

Example 16. The radii of the two ends of a pail of height 24 cm are 15 cm and 5 cm. Find the volume of the pail.

Solution:

Given:

The radii of the two ends of a pail of height 24 cm are 15 cm and 5 cm.

The volume of the pail = (Volume of the conical OAB) – (Volume of. the cone OCD).

Now, ΔOPB ~ ΔOQD,

∴ \(\frac{\mathrm{PO}}{\mathrm{OQ}}=\frac{\mathrm{PB}}{\mathrm{QD}}\)

⇒ \(\frac{h_1}{h_2}=\frac{15}{5}=3\)

⇒ h₁ = 3h₂

Again, PQ = OP – OQ

⇒ h₁ -h₂ = 24

⇒ 3h₁ – h₂ = 24

⇒ 2h₂ = 24

⇒ h₂ = \(\frac{24}{2}\)

⇒ h₂ = 12

∴ h₁ = 3h₂ = 3 x 12 = 36

So, the volume of the cone OAB = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 15² x 36 cc

= \(\frac{22}{7}\) x 5 x 15 x 36 cc

and the volume of the cone OCD = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 15² x 12 cc

= \(\frac{22}{7}\) x 5 x 5 x 4 cc

∴ the volume of the pail = \(\left(\frac{22}{7} \times 5 \times 15 \times 36-\frac{22}{7} \times 5 \times 5 \times 4\right) \mathrm{cc}\)

= \(\frac{22}{7} \times 5(15 \times 36-5 \times 4) \mathrm{cc}\)

= \(\frac{22}{7} \times 5(540-20) \mathrm{cc}=\frac{22}{7} \times 5 \times 520 \mathrm{cc}\)

= \(\frac{57200}{7} \mathrm{cc}=817.1 \frac{3}{7} \mathrm{cc}\)

Hence the volume of the pail= 8171 \(\frac{3}{7} \mathrm{cc}\) cc.

Example 17. A hemisphere of internal diameter 36 cm is filled with water. If this water is poured into right circular bottles of radius 3 cm and of height 6 cm, then how many bottles will be needed?

Solution:

Given:

A hemisphere of internal diameter 36 cm is filled with water. If this water is poured into right circular bottles of radius 3 cm and of height 6 cm

The radius of the hemisphere = \(\frac{36}{2}\)cm = 18 cm

∴ the volume of the hemisphere = \(\frac{2}{3}\) x π x 18³ cc

The radius of each right circular bottle = 3 cm and its height = 6 cm.

∴ The volume of each bottle = π x 3² x 6 cc

Let the number of bottles to be needed be x.

∴ x x π x 3² x 6 = \(\frac{2}{3}\) x π x 18³.

⇒ x = \(\frac{2 \times 18 \times 18 \times 18}{3 \times 3 \times 3 \times 6}\)

⇒ x = 72

Hence the required number of bottles = 72.

“Volume and surface area of different solids examples Class 10”

Example 18. The ratio of the volumes of a cylinder and a sphere is 1 : 3 and the ratio of the radii of the base of the cylinder and of the sphere is 1 : 3. If the sum of the height and radius of base of the cylinder is 78 cm, then what will be the height of the cylinder?

Solution:

Given:

The ratio of the volumes of a cylinder and a sphere is 1 : 3 and the ratio of the radii of the base of the cylinder and of the sphere is 1 : 3. If the sum of the height and radius of base of the cylinder is 78 cm

Let the radii of the cylinder and the sphere be x cm and 3x cm respectively.

Also let the volumes of the cylinder and the sphere be V and 3V respectively.

As per question, x + h = 78 ………..(1)

Again, V = πx²h cu-unit and 3V = \(\frac{4}{3}\)π x (3x)³ cu-unit

⇒ 3πx²h = 36πr³ ⇒ h = \(\frac{36 x^3}{3 x^2}\) = 12x

∴ from (1) we get, 12x + x = 78

or, 13x = 78

or, x = \(\frac{78}{13}\) = 6

and h = 12x

⇒ h = 12 x 6 = 72.

Hence the height of the cylinder = 72 cm.

Example 19. The total surface area of a solid hemisphere is 1848 sq-cm. A solid right circular cone is made by melting this hemisphere. If the radius of base of the cone is equal to the radius of the hemisphere, then what will be the height of the cone?

Solution:

Given:

The total surface area of a solid hemisphere is 1848 sq-cm. A solid right circular cone is made by melting this hemisphere. If the radius of base of the cone is equal to the radius of the hemisphere

Let the radius of the hemisphere be r cm.

So, the total surface area of the hemisphere = 3πr²sq-cm

As per the question, 3πr² = 1848

or, 3 x \(\frac{22}{7}\) x r² = 1848

or, r² = \(\frac{1848 \times 7}{3 \times 22}\)

or, r² = 196

Also, the volume of the hemisphere = \(\frac{2}{3}\)πr³ cc

and the volume of the cone = \(\frac{1}{3}\)πr²hcc

As per question, \(\frac{2}{3}\)πr³ = \(\frac{1}{3}\)πr²h

⇒ h = \(\frac{2 \pi r^3}{\pi r^2}\)

⇒ h = 2r

⇒ h = 2 x 14

⇒ h = 28

Hence the height of the cone = 28 cm.

Example 20. Half of a tank of length 5 metre, breadth 4 metre and height 2.2 metre, is filled with water. How many iron-bullets of radius 5 cm each should be completely immersed into the water of the tank so that the tank will full to the brim?

Solution:

Given:

Half of a tank of length 5 metre, breadth 4 metre and height 2.2 metre, is filled with water.

The radius of each iron bullet = 5 cm

∴ the volume of each iron-bullet = \(\frac{4}{3}\)π x 5³ cc

Let the required number of iron-bullets be x.

the total volume of x bullets = x x \(\frac{4}{3}\)π x 5³ cc .

Since half of the tank is filled with water, the Water-level of the tank will be increased \(\frac{2.2}{2}\) m = 1.1 m to be filled to the brim.

So, the volume of the raised-water = 5 x 4 x 1.1 cubic-metre

= 500 x 400 x 110 cc

As per question, \(\frac{4}{3}\) x \(\frac{22}{7}\) x 5³ x x = 500 x 400 x 110

⇒ x = \(\frac{500 \times 400 \times 110 \times 3 \times 7}{4 \times 22 \times 125}\)

⇒ x = 42000

Hence the required number of iron-bullets = 42000.

Example 21. Half of a tank of length 21 dcm, breadth 11 dcm and depth 6 dcm is full of water. If 100 spheres made of iron and of diameter 21 cm each are fully immersed into this water of tank, then how many dcm of the water-level of the tank will be raised?

Solution:

Given:

Half of a tank of length 21 dcm, breadth 11 dcm and depth 6 dcm is full of water. If 100 spheres made of iron and of diameter 21 cm each are fully immersed into this water of tank

The length of the tank = 21 dcm

breadth of the tank =11 dcm; and

depth of the tank = 6 dcm.

Let the water-level of the tank rises x dcm after the complete immersion of 100 iron-spheres.

∴ the volume of the raised-water = (21 x 11 x x) cubic-dcm.

Again, diameter of each sphere = 21 cm

∴ radius of each sphere = \(\frac{21}{2}\) cm = \(\frac{21}{2 \times 10}\) x 10 = 1.05 dcm.

∴ the volume of 100 iron-spheres = 100 x \(\frac{4}{3}\) x \(\frac{22}{7}\) x (1.05)³ cubic-dcm

As per question, 21 x 11 x x = 100 x \(\frac{4}{3}\) x \(\frac{22}{7}\) x (1.05)³

x =\(\frac{100 \times 4 \times 22 \times 1.05 \times 1.05 \times 1.05}{3 \times 7 \times 21 \times 11}\)

= \(\frac{100 \times 4 \times 22 \times 105 \times 105 \times 105}{3 \times 7 \times 21 \times 11 \times 100 \times 100 \times 100}\)

= 2.1

Hence the water-level of the tank will rise 2. 1 dcm.

Example 22. The cross-section of a rectangular parallelopiped wooden log of length 2 metre is squareshaped and each of its side is of length 14 dcm. If the rectangular log is reduced to a right circular log by loosing smallest amount of wood, then how much cubic-metre of wood the log will posses and how cubic-metre of wood will be lost?

Solution:

Given:

The cross-section of a rectangular parallelopiped wooden log of length 2 metre is squareshaped and each of its side is of length 14 dcm. If the rectangular log is reduced to a right circular log by loosing smallest amount of wood

The cross-section of the rectangular parallelopiped wooden log is square-shaped and each of its side is of length 14 dcm.

The rectangular log is reduced to a circular log.

∴ the length of the diameter of the circular area = the length of the side of the square.

∴ the radius ofthe right circular cylindrical wooden

log = \(\frac{14}{2}\) dcm = 7 dcm.

The length of the log = 2 metre = 2 x 10 dcm = 20 dcm

∴ the volume of the rectangular wooden log= 14 x 14 x 20 cubic-dcm

= 3920 cubic-dcm = 3.92 cubic-metre.

Again, the volume of the circular log = \(\frac{22}{7}\) x 7² x 20 cubic-dcm

= 3080 cubic-dcm = 3.08 cubic-metre

∴ the lost wood = (3.92 – 3.08) cubic-metre.

= 0.84 cubic-metre

Hence the right circular cylindrical log will possess 3.08 cubic-metre of wood and the lost wood is 0.84 cubic-metre.

Example 23. What will be the ratio of the volumes of a sphere and a cube if the surface areas of the sphere and the cube are equal?

Solution:

Let the radius of the sphere = r unit

and let the length of each side of the cube = x unit.

As per question, 4πr² = 6x²

⇒ \(\frac{r^2}{x^2}=\frac{3}{2 \pi}\)

⇒ \(\frac{r}{x}=\frac{\sqrt{3}}{\sqrt{2} \cdot \sqrt{\pi}}\)

∴ (volume of the sphere) : (volume of the cube)

= \(\frac{4}{3} \pi r^3: x^3=\frac{\frac{4}{3} \pi r^3}{x^3}=\frac{4}{3} \pi\left(\frac{r}{x}\right)^3\)

= \(\frac{4}{3} \times \pi \times\left(\frac{\sqrt{3}}{\sqrt{2} \cdot \sqrt{\pi}}\right)^3=\frac{4}{3} \times \pi \times \frac{3 \sqrt{3}}{2 \sqrt{2} \pi \sqrt{\pi}}\)

= \(\frac{\sqrt{2} \times \sqrt{3}}{\sqrt{\pi}}=\frac{\sqrt{6}}{\sqrt{\pi}}=\sqrt{6}: \sqrt{\pi}\)

Hence the required ratio =√6 : √π.

Example 24. A right circular cone*is made by cutting a cubical wooden block of side 9 cm each, then what will be the greatest Volume of the cone?

Solution:

Given:

A right circular cone*is made by cutting a cubical wooden block of side 9 cm each

The length of each side of the cube = 9 cm

i.e., if the radius of the required cone be r cm, then 2r = 9 ⇒ r = \(\frac{9}{2}\)

∴ the height of the cone = 9 cm

∴ Volume of the cone = \(\frac{1}{3}\)πr²h cc

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x \(\left(\frac{9}{2}\right)^2\) x 9 cc

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x \(\frac{9}{2}\) x \(\frac{9}{2}\) x 9 cc

= \(\frac{2673}{14}\) cc

= 190.93 cc (approx.)

Hence the greatest volume of the cone = 190.93 cc (approx.)